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Eigenvalues and Eigenfunctions in S-L Problems 📂Lebesgue Spaces

Eigenvalues and Eigenfunctions in S-L Problems

Definition1

If the Sturm-Liouville differential equation

[p(x)u(x)]+[q(x)+λw(x)]u(x)=0 \begin{equation} \left[ p(x)u^{\prime}(x) \right]^{\prime}+\left[ q(x) +\lambda w(x) \right]u(x)=0 \end{equation}

has a solution uLr2(a,b)u \in L_{r}^{2}(a,b) different from 00, then λ\lambda is called an eigenvalue, and the corresponding uu is referred to as the eigenfunction.

Explanation

Let’s assume the weighting function is w(x)=1w(x)=1. Then, (1)(1) can be written as follows.

p(x)u(x)+p(x)u(x)+q(x)u(x)+λu(x)= 0    p(x)u(x)p(x)u(x)q(x)= λu(x) \begin{equation} \begin{aligned} && p(x)u^{\prime \prime}(x) +p^{\prime}(x)u^{\prime}(x)+q(x)u(x)+\lambda u(x) =&\ 0 \\ \implies && -p(x)u^{\prime \prime}(x) -p^{\prime}(x)u^{\prime}(x)-q(x) =&\ \lambda u(x) \end{aligned} \end{equation}

Let’s suppose the operator D:C2[a,b]C[a,b]D:C^{2}[a,b] \to C[a,b] is as follows.

Du(x):=p(x)d2u(x)dx2p(x)du(x)dxq(x)u(x) Du(x):=-p(x)\frac{ d ^{2}u(x)}{ d x^{2} }-p^{\prime}(x)\frac{ d u(x)}{ d x }-q(x)u(x)

Then, (2)(2) can be expressed as below.

Du=λu Du =\lambda u

In the S-L problem, λ\lambda becomes the eigenvalue, and uu is the corresponding eigenfunction.


  1. Ole Christensen, Functions, Spaces, and Expansions: Mathematical Tools in Physics and Engineering (2010), p218 ↩︎